show that $a(1-b)$ and $b(1-a)$ can not both be greater than 1/4 for any two $a,b \in \mathbb{N}$

Question

Please help me solve this problem: show that $a(1-b)$ and $b(1-a)$ can not both be greater than 1/4 for any two $a,b \in \mathbb{N}$.

Answer 1

Their sum is $a+b-2ab=1/2-2(a-1/2)(b-1/2)$. Any constraints that ensure $(a-1/2)(b-1/2)\ge 0$ complete the proof, since $a+b-2ab\le 1/2$ contradicts $a(1-b),\,b(1-a)$ both exceeding $1/4$. In the example at hand, $a,\,b$ are natural so are both $\ge 1$, which is sufficient.

Answer 2

If $a$ and $b$ are allowed to be nonnegative real, this still holds.

We may assume that both $a$ and $b$ satisfy $a,b \le 1$ and in fact $a,b >0$, lest one of $a(1-b)$, $b(1-a)$ is negative or 0. Let us assume WLOG that $a$ and $b$ satisfy $a \geq b$. Then $a(1-a)$ is no greater than $\frac{1}{4}$ for all $a \in \mathbb{R}$, which implies that $b(1-a)$ is no greater than $\frac{1}{4}$ (because $b \le a$ and $a \le 1$ and $1-a \ge 0$).